Boost::spirit::qi - 如何构建一个设置 属性 的解析规则？

Question

我想构建一个规则，从解析的行中获取一些参数，然后将一些参数设置为常量。那可能吗？下面是一个（无效的）示例来说明我正在尝试做的事情。我想我在这里使用 _r1 不正确，但我不确定如何做正确的事情。假设我不想在将 r 发送到解析器之前只修改它。

#include <boost/spirit/include/qi.hpp>
#include <boost/spirit/include/qi_plus.hpp>
#include <boost/spirit/include/qi_sequence.hpp>
#include <boost/spirit/include/qi_string.hpp>
#include <boost/spirit/include/phoenix_core.hpp>
#include <boost/spirit/include/phoenix_operator.hpp>
#include <boost/spirit/include/phoenix_object.hpp>
#include <boost/fusion/include/adapt_struct.hpp>
#include <boost/fusion/adapted/struct/adapt_struct.hpp>
#include <boost/phoenix/bind/bind_function.hpp>
#include <string>

using namespace boost::spirit::qi;
struct Sample
{
    int a;
    int b;
};
BOOST_FUSION_ADAPT_STRUCT(Sample, a , b)

const rule<std::string::const_iterator, Sample()> AnythingAndOne = int_ >> eps[_r1.b = 1] >> eoi;


int main()
{
    std::string z("3");
    Sample r;
    parse(z.begin(), z.end(), AnythingAndOne, r);
    return 0;
}

Answer 1

同样，参考 Boost Spirit: "Semantic actions are evil"? 我会避免语义操作。

您可以使用qi::attr:

直接合成一个特定的属性值

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#include <boost/spirit/include/qi.hpp>
#include <boost/fusion/include/adapt_struct.hpp>
#include <boost/fusion/include/io.hpp>

struct Sample {
    int a;
    int b;
};
BOOST_FUSION_ADAPT_STRUCT(Sample, a , b)

namespace qi = boost::spirit::qi;

int main()
{
    std::string const z("3");
    Sample r;
    qi::rule<std::string::const_iterator, Sample()> AnythingAndOne
         = qi::int_ >> qi::attr(1) >> qi::eoi;

    if (parse(z.begin(), z.end(), AnythingAndOne, r))
        std::cout << "Parsed: " << boost::fusion::as_vector(r) << "\n";
    else
        std::cout << "Parse failed\n";
}

版画

Parsed: (3 1)